To determine the percentage of iron present as Fe(III) in the sample of wustite (Fe_0.93O_1.00), we need to assume that the rest of the iron that is not Fe(II) is present as Fe(III). This is because wustite, which nominally has the formula FeO, has a non-stoichiometric composition due to the presence of both Fe(II) and Fe(III) ions that maintain charge neutrality in the lattice.

Given the formula Fe_0.93O_1.00, for every one oxygen atom, there are 0.93 irons. In terms of charge, the oxygen anion has a charge of -2. The formula can be interpreted as having 0.93 moles of Fe per mole of wustite, and we need to balance the charges considering the presence of both Fe(II) and Fe(III).

Fe(II) has a charge of +2, and Fe(III) has a charge of +3. If we let x be the fraction of Fe(III), then the fraction of Fe(II) would be 0.93 - x because the total amount of iron is 0.93 moles of iron (Fe). Now, we can set up a charge balance equation:

Charge from Fe(II) + Charge from Fe(III) = Charge from O

(0.93 - x)(+2) + (x)(+3) = 1(+2)

2(0.93 - x) + 3x = 2

1.86 - 2x + 3x = 2

Combining like terms gives:

1.86 + x = 2

x = 2 - 1.86

x = 0.14

So, 0.14 moles of Fe are in the form of Fe(III) per mole of wustite.

To find the percentage of Fe(III), you divide the number of moles of Fe(III) by the total moles of Fe and multiply by 100%:

$$ \text{Percentage of Fe(III)} = \left( \frac{x}{0.93} \right) \times 100\% $$

$$ \text{Percentage of Fe(III)} = \left( \frac{0.14}{0.93} \right) \times 100\% $$

$$ \text{Percentage of Fe(III)} = (0.1505) \times 100\% $$

$$ \text{Percentage of Fe(III)} = 15.05\% $$

Therefore, 15.05% of the iron is present in the form of Fe(III) in the sample of wustite Fe_0.93O_1.00.